Question

The mean scores of a random sample of 17 students who took a special test is 83.5. If the standard deviation of the scores is 4.1, and the sample comes from an approximately normal population, find the point and the interval estimates of the population mean adopting a confidence level of 95%. With explanation

Answer 1

The point estimate is the mean scores of a random sample of 17 students who took a special test: $\mu=83.5.$

The critical value for $\alpha = 0.05$ and $df = n-1 = 16$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.119905$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times \dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times \dfrac{s}{\sqrt{n}})$$ $$=(83.5-2.119905\times \dfrac{4.1}{\sqrt{17}},$$ $$83.5-2.119905\times \dfrac{4.1}{\sqrt{17}})$$ $$=(81.392,85.608)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $81.392 < \mu < 85.608,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(81.392, 85.608).$

The interval estimate of the population mean adopting a confidence level of 95% is $81.392 < \mu < 85.608.$